4

If you live in a two-dimensional world, would it be possible to explain the angular momentum?

Qmechanic
  • 201,751
Syeda Rabita Alam 1821333042
  • 41

5 Answers5

5

Angular momentum of a point mass is defined as $\mathbf p=m(\mathbf r\times \mathbf v) $. In the 2D case, the components of $\mathbf r$ and $\mathbf v$ along the $z$ direction will be zero. Thus,

$$\mathbf p=m((x\mathbf{\hat i}+y\mathbf {\hat j})\times(v_x\mathbf{\hat i}+v_y\mathbf{\hat j}))=\boxed{m(x v_y-y v_x)\mathbf{\hat k}}$$

Now, you might think that the angular momentum doesn't exist because of the direction of the angular momentum vector that you obtain ($\mathbf{\hat k}$). However, it is important to remember that this direction doesn't really have a strong physical implication. The direction obtained by the cross product is only used to define the sense of motion (clockwise or anti-clockwise). Thus the direction is just a mathematical result and doesn't restrict us from computing the angular momentum as long as we can figure out the sense in which the body is moving.

  • May I know the reason for the downvote? –  Jun 01 '20 at 11:01
  • I have a point to clarify. That a circle can be made on a 2D plane is a good enough reason for an angular momentum to be defined. However, the basis vector for such a vector space would have Dim $\mathcal{V}$=2.Since, the cross product defines a vector perpendicular to both the co-planar vectors; wouldn't the new vector be defining a normal to the plane vector with its own vector space? – Ishika_96_sparkle Jun 01 '20 at 11:43
  • @Ishika_96_sparkle As I said, the direction of the angular momentum doesn't really have any physical significance. It can be seen as a mathematical trick to evade defining the plane and the sense of the motion of any particle. We can conveniently use the plane of motion and the sense of motion instead of the direction of angular momentum vector without any loss in the information of any scenario. –  Jun 01 '20 at 12:44
  • If, as you wrote that the cross-product just fixes the sense of direction only and no physical significance, then how will you define $\vec{\nabla} \times \vec{B}$ or $\vec{\nabla} \times \vec{E}$ in a two dimensional world? For instance, the effect of current carrying loop? – Ishika_96_sparkle Jun 01 '20 at 13:42
  • 2
    Hi @Ishika_96_sparkle, the ideas of curls (like cross-products) are not easily generalised to spaces of different dimension, as the other answers have mentioned as well. These objects are replaced by more fundamental ones. For example, in the case of Maxwell's Equations, the form that we are used to is useful in 3D. In 2D EM the magnetic field only has 1 component. It's not a vector, it's a scalar. but if you realise that $\vec{E} $ and $\vec{B}$ are actually part of a more fundamental quantity called the Electromagnetic Field Tensor $F_{\mu\nu}$, this easily generalises to other dimensions. – Philip Jun 01 '20 at 14:38
  • Thanks @Philip. I don't think i have fully grasped the idea of EM tensors from the Griffiths book. So, can't argue about that. I had imagined a scenario in which a 2D being makes a wire loop and switches on the current. This would be her solenoid. In which direction would the $\vec{B}$ point if the current circulates in the anti-clockwise/clockwise direction? My contention is that we cannot simply take a vector as a purely directional thing. There is a physical effect also because of the magnitude of the vector field at a point. – Ishika_96_sparkle Jun 01 '20 at 15:36
  • 1
    @Ishika_96_sparkle You are right when it comes to magnetic field, but in the case of angular momentum, if a 2D creature saw another 2D creature rotating in a circle around it, then nothing stops that creature from writing out the another creature's angular momentum as $mvr$. If the other creature starts rotating in the opposite sense, the creature at the center can modify the old angular momentum by adding a negative sign, giving $-mvr$. The inability of using the third dimension doesn't take away their ability of measuring angular momentum (and it's direction in their conventions, not ours). –  Jun 01 '20 at 17:22
2

You have your answers already, I just want to add little more theory.

The cross product defined as vector (or more precisely pseudovector) with magnitude equal to the area of the parallelogram formed by the two multiplied vectors and with direction perpendicular to both of them is applicable only in 3 dimensions.

In 2 dimensions, you can still compute area of the parallelogram, but here assigning some direction makes no sense and resulting value will be scalar (or more precisely pseudoscalar).

In 1 dimension the area is trivially 0, so here, the cross product is always 0.

In more than 3 dimensions there is however a problem. Now, there is infinite many directions that are perpendicular to both multiplied vectors, so which one to pick? This is huge problem for physics, because in special relativity the spacetime and vectors are 4 dimensional. Suddenly, we have no idea how to define angular momentum.

Luckily, mathematicians were able to give sense to cross product in any dimensions. This meaning comes from exterior algebra. In exterior algebra mathematicians defined the operation of wedge product, which creates from two vectors a bivector. This bivector can be identified with antisymmetric tensor of rank 2. If you are not familiar with tensors, you can imagine them as squared matrixes with dimension given by dimension of the space in which you work.

So the cross product of two vectors is more tensor (matrix) than a vector. In 3 dimensions, the antisymmetric tensor has 3 independent nonzero components, which by accident is same as number of components of vector. In 2 dimensions, there is only one independent nonzero component, whereas in 1 dimension the only component is trivially zero (due to antisymmetricity). In 4 dimensions however, you have 6 independent nonzero components and this will no longer fit inside a vector. But the wedge product as defined by exterior algebra is still well defined.

Wedge product has also more general geometric meaning. When you take wedge product of 2 vectors, you get area formed by parallelogram together with some kind of sense of orientation of the parallelogram given by the fact, that wedge product is antisymmetric. That is $v_1 \wedge v_2=-v_2 \wedge v_1$, where $\wedge$ denotes wedge product. So when you exchange the two vectors, the orientation of parallelogram gets reversed. One important point, which was already mentioned in previous answers, is that direction of the vector resulting from cross product is not important. Important thing is only orientation.

When you take wedge product of 3 vectors, you get volume of the parallelogram fromed by this 3 vectors together with sense of orientation, which in case of 3D space means the sense of inside and outside of the volume. When you multiply 4 vectors, you get volume of 4-dimensional parallelogram (which in 3D space is trivially zero, but can be nonzero in 4D space or higher) and so on.

Umaxo
  • 5,818
2

Just adding my two-cents here to the already great answers: angular momentum is not actually a vector even though we're usually taught that it's one. It turns out that -- as a happy accident in 3 dimensions -- it happens to have three independent components that transform under rotations like a vector should.

However, it fails one of the "tests" required by a vector: the mirror test. Imagine an object in 3D rotating the anticlockwise direction in the $xy-$plane in circular motion. It's angular momentum points "along $z$". Now imagine looking at this object in a mirror. You'll see an object rotating clockwise but the angular momentum will still be pointing upwards in the "mirror-world". But you know that an object rotating clockwise must have an angular momentum pointing along $-z$ (i.e. "downwards") and so you could tell the difference between the "real" world and the "mirror" world using this, leading to such objects being called pseudovectors. (These are usually objects created by taking the cross-product of two "proper" or "polar" vectors)

If you find this hard to visualise, another way you could do it is by looking at the magnetic field: it turns out that this quantity also transforms like the angular momentum does. You can read more about this in this appendix.

As mentioned in the answer by Umaxo, this is because these "pseudovectors" aren't actually vectors: they are rather associated with planes. It's a fun exercise in combinatorics to show that in $n$ dimensions, you have $n(n-1)/2$ planes. Immediately, we see why 3 dimensions is special: we have 3 different axes ($x,y,z$), and 3 different planes ($xy,yz,zx$). Angular momentum is associated with rotations, and in general, rotations occur in planes, not about axes, and this explains why we can have angular momentum in 2D because even though there is no "axis" about which to rotate, there is still a plane to rotate in.

Thus in $n$ dimensions, quantities like angular momentum have $n(n-1)/2$ components, and it turns out (though what I've said is hardly a proof) that they are the components of an antisymmetric $n\times n$ quantity called a "tensor" -- which are objects that transform in a specific way, but this is not very important at this level.

So yes, you can define angular momentum in 2D, but it will be a scalar not a vector, just like you can define the magnetic field in 2D which will also be a scalar. The same applies for the magnetic field.

Philip
  • 11,393
0

Yes, it would be possible. Angular momentum would not be a vector (with 3 components) but a scalar (with 1 component). $$L=xp_y-yp_x$$

Thomas Fritsch
  • 39,317
0

Rotational Motion is a 2D motion itself. This motion doesn't care about the existence of a third axis. Just like linear momentum, angular momentum is associated with rotational motion. But, Yeah You wouldn't have like a wobbling, precession and gyroscopic motion which is completely understood in term of angular momentum.

Vector is important when it comes to different directions, but in the case of 2d motion, we would only have one 3d direction (i.e, z-axis, say) which would make all quantity the torque, angular momentum, angular velocity etc scalar(as there is no other direction to worry about).

So yes we would still be able to define the angular momentum even if we would live in a 2D world

The_Lazy_Panda
  • 175
  • 8